Question

Solve x^2 - 3x = -8
Lemme see if I get this right...
x^2 - 3x = -8
(x-1.5)^2 = -8 -1
(x-1.5)^2 = -9
x - 1.5 = +or- sqrt9i
x-1.5 = 3sqrt2?

Answer 1

I am horribly wrong. These are my choices. http://prntscr.com/92694d

Answer 2

You're very close. Give it another go.

x^2 - 3x = -8

x^2 - 3x + (3/2)^2 = -8 + (3/2)^2

(x - (3/2))^2 = -8 + (9/4) = -23/4

x - (3/2) = +/- sqrt(-23)/2

x = (3/2) +/- sqrt(-23)/2

x = (3/2) +/- i*sqrt(23)/2

Answer 3

I don't get what to do.

Answer 4

And the end of your reply is not an answer choice.

Answer 5

@tk

Answer 6

@tkhunny

Answer 7

@pooja195 @kohai

Answer 8

@Kainui @Hero

Answer 9

Your first step needs improvement. You can't just declare it a perfect square, you must make it one. Compare your first step to my fist two steps. Where did you get "-1"?

Answer 10

I got -1.5 by dividing 3x by 2.

Answer 11

You had 3
You divided by 2 to get 3/2.
(3/2)^2 = 9/4
Where did you get -1?

Answer 12

Oh I had a typo there. I meant -1.5 and I guess I stuck with it by accident.

Answer 13

I wanted to do it to both sides :/

Answer 14

What do I do?

Answer 15

It's never negative. You must ADD 9/4 to both sides.

Answer 16

Can ya walk me through the rest? I'm really bad at Algebra II.

Answer 17

Hello?

Answer 18

Already did. It's in my first post.

Answer 19

Is it 3 + or - isqrt23/2?

Answer 20

(3 + or - isqrt23)/2

or

3/2 + or - isqrt23/2?

Take your pick.

Answer 21

Learning just a little LaTeX goes a VERY long way.

\(x = \dfrac{-3\pm i\sqrt{23}}{2}\) or \(x = \dfrac{3}{2}\pm\dfrac{i\sqrt{23}}{2}\)